How do you find the integral of x^5*e^(x^2) ?

Apr 25, 2018

$\frac{1}{2} {x}^{4} {e}^{{x}^{2}} - {x}^{2} {e}^{{x}^{2}} + {e}^{{x}^{2}} + C$

Explanation:

$\int {x}^{5} \cdot {e}^{{x}^{2}} \mathrm{dx}$

$= \frac{1}{2} \int {x}^{4} {e}^{{x}^{2}} \cdot 2 x \mathrm{dx}$

$= \frac{1}{2} \int {x}^{4} {e}^{{x}^{2}} {\mathrm{dx}}^{2}$

Integration by substitution

${x}^{2} = u$

$d \left({x}^{2}\right) = \mathrm{du}$

$= \frac{1}{2} \int {x}^{4} {e}^{{x}^{2}} {\mathrm{dx}}^{2} = \frac{1}{2} \int {u}^{2} {e}^{u} \mathrm{du}$

$= \frac{1}{2} \int {u}^{2} d \left({e}^{u}\right)$

$= \frac{1}{2} \left({u}^{2} {e}^{u} - \int 2 u {e}^{u} \mathrm{du}\right)$

Using integration by Parts again

$= \frac{1}{2} \left({u}^{2} {e}^{u} - 2 \int u d \left({e}^{u}\right)\right)$

$= \frac{1}{2} \left({u}^{2} {e}^{u} - 2 u {e}^{u} - \left(- 2 \int {e}^{u} \mathrm{du}\right)\right)$

$= \frac{1}{2} {u}^{2} {e}^{u} - u {e}^{u} + {e}^{u} + C$

Reverse The Substitution

$= \frac{1}{2} {x}^{4} {e}^{{x}^{2}} - {x}^{2} {e}^{{x}^{2}} + {e}^{{x}^{2}} + C$